Given the equation below: $u(x) =\frac {1}{2a} \int_{x-a}^{x+a} g(\xi) \, d\xi$
With $$u(x) = \begin{cases} 0,& (|x|>\ 1+a) \\ -\frac 1 a (|x|-a-1),& (1-a\le|x|\le1+a) \\ 2, &(|x|<1-a) \end{cases}$$
Find the Fourier transform of $u(x)$ and the Fourier transform of the function below: $$f(x) = \begin{cases} \frac{1}{2a}, &(|x| \le a) \\ 0, &(|x|>a) \end{cases}$$ Then use it to find $g(x)$
I can find the Fourier Transform, but there are two points I am wondering:
1) For the middle part of $u(x)$ $(1-a\le|x|\le1+a)$, can I find the Fourier Transform for $u'(x) = -|x|$ in $(0 \le |x| \le 2a)$ first and then apply Fourier Transform's properties?
2) The Fourier Transform of $f(x)$ is $\frac {\sin\omega a}\omega$, but the scale of $u(x)$ is $(|x| \le 1+a)$, so to find $g(x)$, should I change the Fourier Transform of $f(x)$ to $\frac {\sin\omega (1+a)}\omega$? Because I think the $a$ in $u(x)$ and $f(x)$ are different.